Question 18 John was meeting two friends in Paris, one arriving by train from Munich and the other arriving by train from Barcelona at approximately the same time. Let A and B be the events that the trains are on time. Suppose we know from past experience that, \[ \begin{array}{l} P(A)=0.93 \\ P(B)=0.89 \\ P(A \cap B)=0.87 \end{array} \] Find the probability that train A or train B is on time. 0.87 0.91 0.95 0.83

Step 1 ：Let's denote the event that the train from Munich is on time as A and the event that the train from Barcelona is on time as B. We are given that the probability of A, denoted as P(A), is 0.93 and the probability of B, denoted as P(B), is 0.89. We are also given that the probability of both A and B occurring, denoted as P(A ∩ B), is 0.87.

Step 2 ：We are asked to find the probability that either train A or train B is on time. This is denoted as P(A U B).

Step 3 ：The formula for the probability of either event A or event B occurring is given by P(A U B) = P(A) + P(B) - P(A ∩ B).

Step 4 ：Substituting the given values into the formula, we get P(A U B) = 0.93 + 0.89 - 0.87.

Step 5 ：Calculating the above expression, we find that P(A U B) = 0.95.

Step 6 ：Thus, the probability that either train A or train B is on time is approximately \(\boxed{0.95}\).